Biostat 202B Homework 4

Due May 9, 2024 @ 11:59PM

Problem 1

Answer:

\(L(\theta)=\prod_{i=1}^{n}f(x_i;\theta)=\prod_{i=1}^{n}\frac{e^{-x_i/\theta}}{\theta}=\frac{e^{-\sum x_i/\theta}}{\theta^n}\)

\(l(\theta)=\log L(\theta)=-\sum x_i/\theta-n\log \theta\)

\(\frac{dl(\theta)}{d\theta}=\sum x_i/\theta^2-n/\theta\)

By setting \(\frac{dl(\theta)}{d\theta}=0\), we have \(\hat{\theta}_{MLE}=\frac{\sum x_i}{n}=\bar{x}\)

\(F_X(x)=\int_{0}^{x}\frac{e^{-z/\theta}}{\theta}dz=-e^{-z/\theta}\Big|_{0}^{x}=1-e^{-x/\theta}\) for \(0<x\leq \theta\)

\(P(X\leq 2)=1-e^{-2/\theta}=1-e^{-2/\bar{x}}\)

Problem 2

(a) Answer:

\(L(\theta)=\prod_{i=1}^{n}f(x_i;\theta)=\frac{2^n \prod_{i=1}^{n}x_i}{\theta^{2n}}\) for \(0<x_i\leq \theta\)

\(L(\theta)=\frac{2^n \prod_{i=1}^{n}x_i}{\theta^{2n}}\) for \(0<x_{(1)}< x_{(2)}<...<x_{(n)}\leq \theta\)

\(L(\theta)=\frac{2^n \prod_{i=1}^{n}x_i}{\theta^{2n}}I(0<x_{(1)})...I(x_{(n)}\leq \theta)\)

So, \(\hat{\theta}_{MLE}=x_{(n)}\) is the smallest value that maximizes the likelihood

(b) Answer:

\(F_X(x)=\int_{0}^{x}\frac{2z}{\theta^2}dz=\frac{z^2}{\theta^2}\Big|_{0}^{x}=\frac{x^2}{\theta^2}\) for \(0<x\leq \theta\)

\(F_{X_{(n)}}(x)=P(X_{(n)}<x)=P(X_{1}<x,X_{2}<x,...,X_{n}<x)=P(X_1<x)^n=F_X^n(x)\)

\(E(x_{(n)})=\int_{0}^{\theta}xnf_X(x)F_X^{n-1}(x)dx=\int_{0}^{\theta}xn\frac{2x}{\theta^2}(\frac{x^2}{\theta^2})^{n-1}dx=\frac{2n}{\theta^{2n}}\int_{0}^{\theta}x^{2n}dx=\frac{2n\theta}{2n+1}\neq \theta\)

Hence, \(\hat{\theta}_{MLE}\) is not unbiased

(c) Answer:

The median of distribution means \(F_X(m)=0.5\)

\(\frac{m^2}{\theta^2}=0.5\) \(\Rightarrow m=\frac{\theta}{\sqrt{2}}\)

\(\hat{m}_{MLE}=\frac{\hat{\theta}_{MLE}}{\sqrt{2}}=\frac{x_{(n)}}{\sqrt{2}}\)

Problem 3

Answer:

\(L(\sigma)=\prod_{i=1}^{n}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x_i^2}{2\sigma^2}}\prod_{j=n+1}^{2n}\frac{1}{2\sqrt{\pi}\sigma}e^{-\frac{x_j^2}{4\sigma^2}}\)

If we take log of likelihood, take derivative and set it to 0,

We get \(\hat{\sigma}^2_{MLE}=\frac{2\sum_{i=1}^{n}x_i^2+\sum_{j=i+1}^{2n}x_j^2}{4n}=\frac{1}{2n}(\sum_{i=1}^{n}x_i^2+\frac{1}{2}\sum_{j=i+1}^{2n}x_j^2)\)

Since \(\frac{\sum_{i=1}^{n}x_i^2}{\sigma^2}\sim \chi^2(n)\), \(\frac{\sum_{j=n+1}^{2n}x_i^2}{2\sigma^2}\sim \chi^2(n)\) and \(X_i\) are independent r.v.s, we have

\(\frac{\sum_{i=1}^{n}x_i^2}{\sigma^2}+\frac{\sum_{j=n+1}^{2n}x_i^2}{2\sigma^2}\sim \chi^2(2n)\)

\(\frac{2n\hat{\sigma}^2_{MLE}}{\sigma^2}=\frac{\sum_{i=1}^{n}x_i^2}{\sigma^2}+\frac{\sum_{j=n+1}^{2n}x_i^2}{2\sigma^2}\sim\chi^2(2n)\)

Problem 4

(a) Answer:

\[L(p)=\prod_{i=1}^{n}p(x_i;p)\\ =\prod_{i=1}^{n}p^{x_i}(1-p)^{1-x_i}\\ =p^{\sum_{i=1}^{n}x_i}(1-p)^{n-\sum_{i=1}^{n}x_i}\]

\(l(p)=\log L(\theta)=(\sum_{i=1}^{n}x_i)\log (p)+(n-\sum_{i=1}^{n}x_i) \log (1-p)\)

\(\frac{dl(p)}{dp}=\frac{\sum_{i=1}^{n}x_i}{\theta}-\frac{n-\sum_{i=1}^{n}x_i}{1-\theta}=0\)

\(p=\frac{\sum_{i=1}^{n}x_i}{n}\)

Since \(\theta = \frac{1}{p}\), \(\hat{\theta}_{MLE}=\frac{n}{\sum_{i=1}^{n}x_i}=\frac{n}{n\bar{X}}=\frac{1}{\bar{X}}\)

(b) Answer:

By CLT, \(\sqrt{n}(\bar{X}-p) \xrightarrow{d} N(0,p(1-p))\)

By Delta method,

\(\sqrt{n}(\frac{1}{\bar{X}}-\frac{1}{p})\xrightarrow{d} N(0,\frac{1}{p^4}p(1-p))=N(0,\frac{1-p}{p^3})\)

Problem 5

(a) Answer:

\(L(\theta)=\prod_{i=1}^{n}p(x_i;\theta)=\prod x_i^{3}e^{-x_i/\theta}/(6\theta^{4})\)

\(l(\theta)=\log L(\theta)=\sum_{i=1}^{n}3\log x_i-x_i/\theta-\log 6-4\log \theta\)

\(\frac{dl(\theta)}{d\theta}=\sum_{i=1}^{n}x_i/\theta^2-4/\theta=0\)

\(\hat{\theta}_{MLE}=\frac{\sum_{i=1}^{n}x_i}{4n}=\frac{\bar{x}}{4}\)

(b) Answer:

\(\sqrt{n}(\bar{X}-E(X))=\sqrt{n}(\bar{X}-4\theta)\xrightarrow{d} N(0,4\theta^2)\)

\(\sqrt{n}(\frac{\bar{X}}{4}-\theta)\xrightarrow{d} N(0, \frac{\theta^2}{4})\)

\(\sqrt{n}(\hat{\theta}-\theta)\xrightarrow{d} N(0, \frac{\theta^2}{4})\)

(c) Answer:

\(SE(\hat{\theta})=\sqrt{\frac{Var(\hat{\theta})}{n}}==\sqrt{\frac{\hat{\theta^2}}{4n}}\)

\(CI = (\hat{\theta}-1.96SE(\hat{\theta}),\hat{\theta}+1.96SE(\hat{\theta}))=(\hat{\theta}-1.96\sqrt{\frac{\hat{\theta^2}}{4n}},\hat{\theta}+1.96\sqrt{\frac{\hat{\theta^2}}{4n}})=(\frac{\bar{x}}{4}-1.96\sqrt{\frac{\bar{x}^2}{64n}},\hat{\theta}+1.96\sqrt{\frac{x^2}{64n}})\)